Asked by Omowole

The cost of maintaining a car is partly constant and partly varies with the distance travelled in a given Month.The cost c for a particular month was 1250 when the distance travelled was 300km . In another month,the cost 35000 for a distance of 1200km .
Find:
a) The formula connection c and d
b) The cost of a jounery of 160km
Answered by oobleck
c = a+bd
Now you have
a+300b = 1250
a+1200b = 35000
Solve for a and b, then find a+160b
Answered by peace
,solution
c=a+be
c=1250=a+b×300k
1250=a+300b..........................(1)
3500=a+1200b........................(2)

equation (2)-(1)
3500_1250=a-a+1200b-300b
2250=0+900b
2250=900b
divide both side by 900b
2250\900=900\900=2.5
find the formula connecting c & d
Answered by peace
I just want to get the formula connecting c and d👌
Answered by Blessing
Substitute b=2.5 in equation 1 a+300(2.5) = 1250. a+750 = 1250. Collect like terms. a = 1250-750. a = 500.
Answered by Blessing
Connecting formula = c = 500+2.5d
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